Consider two metal blocks, one at 500K and another at 300K, they are bought to contact with each other until they reach a common temperature. Find the entropy change of each object and the total change considering there is no change in the volume of the metal blocks in the process and no the other interactions excluding the heat transfer between the two blocks.Also is this process irreversible, why?

also $C_v= 10 J/K$ for both the metal blocks

My attempt: This process is irreversible, as once it reaches the equilibrium temperature it cannot go back to the initial state without external intervention that is to say heat wont flow from cold body to hot body and also when two bodies are at the same temperature heat doesn't flow from one to the other to increase the temperature of one and to decrease the other.

now to compute the $\Delta S$ we can perform the same process very very slowly such that the two systems are always at equilibrium. thus we can write $dS=\frac{\delta q_{rev}}{T}=\frac{dq}{T}=\frac{C_vdT}{T}$ thus for system 1: $\Delta S_1=C_v ln\frac{T_f}{T_i}$ and thus $\Delta S_{TOTAL}=C_v ln\frac{T_{f}}{T_{i1}}+C_v ln\frac{T_{f}}{T_{i2}}$

My questions : I have a feeling my reasoning is not rigorous enough and my definitions are going in circles . and my analytic solution is wrong as the i'm basically using $\delta q$ of the irreversible process and using the same information in the assumed reversible process.


1 Answer 1


This answer is correct. After determining the final temperature Tf, you have gone back to the initial state of each block and, to get its change in entropy, you have subjected it to an alternate reversible process in which, instead of contacting it with the other block, you have contacted it with an infinite continuous sequence of constant temperature reservoirs running from its initial temperature to its final temperature. At least, that is what your equation implies. This is definitely the correct way to determine the entropy change.

  • $\begingroup$ The difference between the irreversible process and the reversible process is in the surrounding. For the former, there is no surrounding, and the entropy of the universe increases. For the latter, the surrounding has a net decrease in entropy (you are heating up the hot reservoirs while cooling down the cold reservoirs), offsetting the net increase of the two blocks such that the entropy of the universe stays constant. $\endgroup$
    – Karsten
    Aug 27, 2019 at 18:19
  • $\begingroup$ @KarstenTheis. With all respect, my take on this is entirely different. In my judgment, there is no need to consider the surroundings at all. My focus, instead, is on the two blocks. I put each of them separately through an alternative reversible process between the same two end states as the irreversible process. The difference between the reversible and irreversible processes is that, in the irreversible process, all the heat transfer takes place at the intermediate interface temperature of 400 K while in the reversible process, this is not the case. $\endgroup$ Aug 27, 2019 at 19:15
  • $\begingroup$ I agree with you, there is no need to consider the surrounding to get the correct answer. I was trying to address the OPs struggle to see why it is fine to run a reversible process to get an answer for an irreversible process. It is kind of surprising because for the reversible process, there is no overall increase in entropy, yet the result for the blocks (increase in entropy) is correct. Does that make sense? $\endgroup$
    – Karsten
    Aug 27, 2019 at 19:28
  • $\begingroup$ @KarstenTheis. Yes, it does make sense. In the pair of reversible paths, net entropy is transferred from the surroundings (reservoir sequences) to the blocks. This makes up for the entropy that was generated within each of the blocks in the irreversible process. $\endgroup$ Aug 27, 2019 at 19:52
  • $\begingroup$ Here is a primer containing a simple cookbook recipe for determining the change in entropy for any irreversible process: physicsforums.com/insights/grandpa-chets-entropy-recipe The article also includes worked examples. Enjoy!!! $\endgroup$ Aug 28, 2019 at 0:14

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